Line Integral over circle region

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Homework Statement


Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.

Homework Equations




The Attempt at a Solution


parametrise
x=1/2cos(t)
y=1/2sin(t)
0≤t≤2π
ds=√dx2+dy2
=√(1/2)2-sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2(sin2(t)+cos2(t))
ds=1/2

∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π

I evaluated this integral and got 0. Is this because C is a simple closed curve?
 

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  • #2
Ray Vickson
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Homework Statement


Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.

Homework Equations




The Attempt at a Solution


parametrise
x=1/2cos(t)
y=1/2sin(t)
0≤t≤2π
ds=√dx2+dy2
=√(1/2)2-sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2(sin2(t)+cos2(t))
ds=1/2

∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π

I evaluated this integral and got 0. Is this because C is a simple closed curve?
I evaluated the (correct) integral and obtained an answer of ##\pi/n##, where ##n## is an integer whose exact value I will not divulge at this time. I assumed the circle was traversed in a counter-clockwise direction, which you did not mention in the problem statement.
 
  • #3
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12
The problem statement didn't mention the direction. So I just assume anti-clockwise.

Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.
 
  • #4
Ray Vickson
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The problem statement didn't mention the direction. So I just assume anti-clockwise.

Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.
PF rules forbid me from telling you exactly where your mistake lies. All I can do is urge you to be more careful, and to read the question carefully.
 
  • #5
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12
Argh!

The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.
 
  • #6
SteamKing
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Argh!

The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.
Since C is a circle, perhaps a more natural expression for ds would be r ⋅ dθ
 
  • #7
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12
I found a source online that suggested if the circle is not at the origin then:
x = h + r cos(t)
y = k + r sin(t)
so
x = 1/2 + (1/2 cos(t))
y = 1/2 sin(t)
edit: FOUND THE INFO I WAS LOOKING FOR!

Circle with Center at Point (h,k)
http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
where the center is (h,k)
and the radius is r

so

x= 1/2cos(t)-1/2
y= 1/2sin(t)
 
  • #8
Ray Vickson
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I found a source online that suggested if the circle is not at the origin then:
x = h + r cos(t)
y = k + r sin(t)
so
x = 1/2 + (1/2 cos(t))
y = 1/2 sin(t)
edit: FOUND THE INFO I WAS LOOKING FOR!

Circle with Center at Point (h,k)
http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
where the center is (h,k)
and the radius is r

so

x= 1/2cos(t)-1/2
y= 1/2sin(t)
In one place you wrote ##x = (1/2) + (1/2)\cos(t)##, while in another place you wrote ##x = -(1/2) + (1/2)\cos(t)##. So, which is it?
 

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